5
$\begingroup$

In a recent question, I asked wether well-pointedness was indeed necessary to make the canonical inclusion $$X\hookrightarrow M_f^{\bullet}$$ a pointed cofibration. Here $X,Y$ are pointed topological spaces (compactly generated weak Hausdorff spaces to be precise), $f:X\to Y$ is a pointed map and $M_f^{\bullet}$ its pointed mapping cylinder. Unless I am deeply mistaken, the canonical inclusion is always a pointed cofibration (and the proof is virtually identical to the unpointed case), however, my main source on this material, Jeff Strom's Modern Classical Homotopy Theory, tells me that one should make the additional assumption of $Y$ being well-pointed, that is, the inclusion of the base point $\ast\hookrightarrow Y$ is an unpointed cofibration.

I could imagine well-pointedness of $Y$ being useful if we wanted to show $X\hookrightarrow M_f^{\bullet}$ to be an unpointed cofibration, although I couldn't show it actually is one. I've checked J.P. May's account on the subject in his A concise Course in Algebraic Topology, and well-pointedness doens't seem to play nearly as big a role as in Strom's book, at least May puts less emphasis on this condition.

What's the deal with well-pointedness? Am I wrong about $X\hookrightarrow M_f^{\bullet}$ necessarily being a pointed cofibration? Why would I need well-pointed spaces, when in the pointed context I am only interested in extending pointed homotopies? Lastly, is it true that when $Y$ is well pointed, the canonical inclusion $X\hookrightarrow M_f^{\bullet}$ is an unpointed cofibration?

I am aware of a theorem that states that if $(\ast\in )A\rightarrow X$ is a pointed map and $A$ is well-pointed, then $A\rightarrow X$ is a pointed cofibration iff it is an unpointed cofibration. This seems like one important benefit of well-pointedness, but it doen't relate to my problem at hand.

EDIT @JustinYoung Thank you for your patience with this. I am not convinced by your counter example, here's why: the pointed mapping cylinder of the identity of $X$ should be homeomorphic to the countable union of closed segments with extremities $(1/n,0,0)$ and $(1/n,1/n,0)$ plus the origin as a subspace of $\Bbb R^3$ (and also to $X\rtimes I$). Thus, $M^{\bullet}_{id}\rtimes I$ should be homeomorphic to the countable union of full squares with corners $(1/n,0,0),(1/n,1/n,0),(1/n,1/n,1/n),(1/n,0,1/n)$ plus the origin as a subspace of $\Bbb R^3$ (and also to $(X\rtimes I)\rtimes I$). In this picture, $X\rtimes I\cup_X M_{id}^{\bullet}\rtimes\{0\}$ is all that has either $x=y$ or $z=0$. If you compress all these squares simultaneously onto their bottom right corner and sides, you get a strong deformation retraction of $M^{\bullet}_{id}\rtimes I$ onto $X\rtimes I\cup_X M_{id}^{\bullet}\rtimes\{0\}$. If you want I can can add my proof of the fact that the canonical inclusion $X\hookrightarrow M_f^{\bullet}$ is always a pointed cofibration, regardless of wether $Y$ is well pointed or not, and you can tell me if you agree or disagree. I can't to it tonight, but I might get to it tomorow.

$\endgroup$
9
  • $\begingroup$ As for the first question, let $X$ be the basepoint of $Y$ and let $f$ be the inclusion $X = * \to Y$. Then, the mapping cylinder $M_f = Y$ and the inclusion $X \to M_f$ is just the inclusion $X = * \to Y$. This is clearly a cofibration if and only if the base point is nondegenerate. $\endgroup$ Jul 26 '12 at 22:23
  • $\begingroup$ @JustinYoung Thanks for the comment. I'm not sure which question you answer though, what question is this an answer to? $\endgroup$ Jul 26 '12 at 22:58
  • $\begingroup$ The question is whether $X \to M_f$ is a pointed fibration, when $Y$ is not well pointed. $\endgroup$ Jul 27 '12 at 0:35
  • $\begingroup$ @JustinYoung I'm pretty sure it is a pointed cofibration regardless of wether $Y$ is well pointed or not: the definitions of Strom and May asks for the extendability of homotopies in the pointed category: during the whole process, the base point is always sent to the base point of the target space, so we don't need well pointedness of $Y$ as the base point is not allowed free roam. $\endgroup$ Jul 27 '12 at 1:15
  • $\begingroup$ Consider the example $X = \{0\}\cup \{1/n \mid n>0\}$, with basepoint $0$, and form the mapping cylinder of $id: X \to X$. Then, you can show that $X \to M_{id}$ is not a pointed cofibration by showing that $M_{id} \rtimes I$ does not retract onto $X\rtimes I \cup_X M_{id} \rtimes \{0\}$. $\endgroup$ Jul 31 '12 at 2:03
1
$\begingroup$

Here are some thoughts towards an answer.

Remark 1. In Tammo tom Dieck's Algebraic Topology, Section 5.3 Problem 2 states what you want, namely that the inclusion $X \to M_f^{\bullet}$ of the "top" of the reduced mapping cylinder is a based cofibration. I'm not sure if tom Dieck has some mild point-set assumptions in the background.

Remark 2. A based map $i \colon A \to X$ is a based cofibration if and only if the "hat-shaped" map $$X \cup_A (A \wedge I_+) \to X \wedge I_+$$ admits a retraction. Here $$X \wedge I_+= X \times I / ({\ast} \times I) = X \rtimes I$$ denotes the reduced cylinder on $X$. That part of the argument is formal, like in the unbased case.

Remark 3. As a warm-up, we can focus on the case $f = \mathrm{id}_X$, as in your example above. In other words, is the inclusion $X \to X \wedge I_+$ of the "top" of the reduced cylinder a based cofibration?

If $X$ is exponentiable (i.e. the functor $X \wedge - \colon \mathrm{Top}_* \to \mathrm{Top}_*$ admits a right adjoint), then $X \to X \wedge I_+$ is a based cofibration. This follows from Remark 2 and the fact that the endpoint inclusion $\{ 1 \} \to I$ is a cofibration.

I don't see immediately if the answer is always yes, without assumption on the based space $X$.

$\endgroup$
3
  • $\begingroup$ Thank you for your answer. In the category of compactly generated weak Hausdorff spaces, all objects are exponentiable, so I guess we are in agrement. $\endgroup$ Mar 29 '18 at 21:08
  • $\begingroup$ That's right. If you're working in a convenient category of spaces $S$, then every space is exponentiable in the unbased sense (i.e., the functor $X \times - \colon S \to S$ has a right adjoint). It follows that every based (convenient) space is exponentiable in the based sense (see ncatlab.org/nlab/show/pointed+object#ClosedMonoidalStructure). Remark 3 then answers the question in the affirmative. There still remains the question about naive topological spaces. I'm not sure if we need assumptions on the based space $X$ then. $\endgroup$ Mar 30 '18 at 2:03
  • $\begingroup$ More precisely, Remark 3 takes care of the case $f = \mathrm{id}_X$, and I think the argument can be adapted to any based map $f \colon X \to Y$. $\endgroup$ Mar 30 '18 at 2:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.